User talk:Chessfan/work1

starting Chessfan (talk) 16:38, 3 May 2010 (UTC)

Speaking about the $$A B C u$$ formula, I think it would be only confusing to try to explain it by some passive rotations. I still have not understood what was meant by extrinsic rotations in the until now existing text. Chessfan (talk) 18:26, 3 May 2010 (UTC)

I had the temptation to introduce some geometric algebra formulations. But there are so few people learning it ! It would be very interesting because the Rotors which replace the matrices are much more explicit : they give the angle and the axis, and are coordinate-free. The interpretation of relations is geometrically very concrete ; there is no puzzling matrix order. Chessfan (talk) 18:36, 3 May 2010 (UTC)

I must even confess that my interest for GA was the main reason pushing me to take a look at Euler rotations. By the way I am still a very young student (19...... *4).Chessfan (talk) 18:41, 3 May 2010 (UTC)

One must also notice that the (z-x-z)matrice in the Table of matrices is not the same as the matrice given in Matrix notation, but coincides with ours if we switch $$\alpha$$ and $$\gamma$$. There is some work to do if we want to put the table in the article. Chessfan (talk) 07:18, 4 May 2010 (UTC)

You are free to use my work in a new article. Feel free to correct my English, but if you want to modify the maths please tell me before ! Chessfan (talk) 07:42, 4 May 2010 (UTC)

I reintroduced reluctantly the mention of quaternions, suppressing only the too sophisticated mention of special unitary group. Why reluctantly ? Because quaternions are an even subalgebra of GA (geometric algebra) which is much easier to manipulate and interprete. Speaking about quaternions without mentioning GA is a shame, but alas corresponds to the present diffusion of knowledge !Chessfan (talk) 16:56, 4 May 2010 (UTC)

After taking a closer look at the extrinsic rotation paragraph I come finally to the conclusion that it coincides with my demonstration of the $$ABCu$$ formula. What is disturbing is the fact that it is expressed in an operator writing instead of matrix algebra, thus reversing the formulas (life is not simple ...). The question is now : is it useful to introduce the appellations ''intrinsic, extrinsic ? '' —Preceding unsigned comment added by Chessfan (talk • contribs) 17:19, 4 May 2010 (UTC) Sorry I change my mind ; it is inextricable because in the intrinsic and extrinsic sections the (XYZ) and (xyz) references are interchanged with the initial definitions ! We should forget those sections. Chessfan (talk) 18:31, 4 May 2010 (UTC)Chessfan (talk) 18:35, 4 May 2010 (UTC)

Perhaps a last try. The main interpretation of Euler rotations can be called intrinsic because the successive rotation axis move with the body ; the second interpretation can be called extrinsic because the successive axis do not move, they are exterior to the body. That means that indeed an extrinsic rotation has nothing to do with what we call a passive transformation.Chessfan (talk) 21:09, 4 May 2010 (UTC)

One must learn by ones errors. There is indeed some relationship between these notions and passive rotations. Look again at relation : $$(t)_e=A B C (u)_e$$

We can write it : $$(u)_e=C^T B^T A^T (t)_e=C(-\gamma) B(-\beta) A(-\alpha)(t)_e$$

That means that $$(u)_e$$ represents the coordinates in the backward rotated initial body reference basis (which now remains fixed), of a vector $$t$$ of the body.

If we write $$A'(\alpha')=A(-\alpha), etc ...$$ we get : $$(u)_e=C' B' A' (t)_e$$ where the $$A'(\alpha')$$ matrices have the same structure than the $$A(\alpha)$$matrices, but are written in the reversed order.

I think there is no need to introduce that in the article, because in passive rotations the idea of Euler rotations no more appears.Chessfan (talk) 09:50, 6 May 2010 (UTC)

intrinsic and extrinsic

 * The main interpretation of Euler rotations can be called intrinsic because the successive rotation axis move with the body ; the second interpretation can be called extrinsic because the successive axis do not move, they are exterior to the body.
 * This is the way I understand it. Is there any other possible meaning?--Guentherwagner (talk) 11:27, 12 May 2010 (UTC)

I agree ; I simply never met those denominations. I am working now on the verification of the angular velocity, and I found, with geometric algebra, that the double formulation of the global rotor, intrinsic and extrinsic, can be completed by a third one, where we simply replace the $$e$$ basis vectors by the corresponding $$h$$ final basis vectors, which is very useful. I am not sure to be able to verify the angular velocity with matrix algebra.Chessfan (talk) 11:55, 12 May 2010 (UTC)

Angular velocity
For the angular velocity I find :


 * $$\omega=(\dot\gamma\sin\beta\sin\alpha+\dot\beta\cos\alpha){\mathbf I}

+(-\dot\gamma\sin\beta\cos\alpha+\dot\beta\sin\alpha){\mathbf J}     +(\dot\gamma\cos\beta+\dot\alpha){\mathbf K},$$

That means a switch between $$\alpha$$ and $$\gamma$$, and a sign change in the middle term. The result is plausible but should be checked with a textbook describing an active rotation ! Chessfan (talk) 16:59, 12 May 2010 (UTC)

I think there is something false in the methodology I employed to establish the above formula. I start again from the roots. Chessfan (talk) 09:24, 13 May 2010 (UTC)

I found what went wrong : mislead by the existing text and happy having established that I could express the global rotor with the $$(h_1,h_2,h_3)$$ vectors I differentiated the rotor expressions, but forgot that the new basis vectors where now variable ! So I switched back to the initial fixed vectors and found easily that in that reference system we get indeed :


 * $$\omega=(\dot\gamma\sin\beta\sin\alpha+\dot\beta\cos\alpha){\mathbf e_1}

+(-\dot\gamma\sin\beta\cos\alpha+\dot\beta\sin\alpha){\mathbf e_2} +(\dot\gamma\cos\beta+\dot\alpha){\mathbf e_3},$$

Switching from $$e$$ to $$h$$ is not simple(I did not try the calculation).

The above expression can of course be established with matrix algebra, but the detailed evaluation seems rather boring and I did not try it. We start by noting that :


 * $$(h_i)=(R^T)(e_i)$$.........$$\dot (h_i)=\dot(R^T)(e_i)$$...... etc ...Chessfan (talk) 14:58, 16 May 2010 (UTC)

A particularly simple check is obtained with the initial instantaneous conditions $$\beta=\gamma=0$$.... $$\dot\alpha=\dot\gamma=0$$. We get :


 * $$\mathbf\omega=(e_1 \cos\mathbf \alpha+e_2 \sin \mathbf\alpha) \dot\mathbf\beta =f_1 \dot\beta$$

which was to be expected. The factor $$\dot\alpha  e_3$$ is also obvious. The other factors can be verified by taking simply.....  $$\dot\alpha=\dot\beta=0$$.

Could somebody check the new formula with a reliable textbook ? Chessfan (talk) 15:14, 16 May 2010 (UTC)

Of course one can write :


 * $$R\omega=(\dot\gamma\sin\beta\sin\alpha+\dot\beta\cos\alpha){\mathbf h_1}

+(-\dot\gamma\sin\beta\cos\alpha+\dot\beta\sin\alpha){\mathbf h_2} +(\dot\gamma\cos\beta+\dot\alpha){\mathbf h_3},$$

but is that useful in Mechanics ? I do not remember. Chessfan (talk) 09:30, 17 May 2010 (UTC)


 * That expression is historically important because Euler solved the equations of motion for the rigid body by changing the reference to the mobile frame. The advantage of this frame is that the inertia tensor is constant and simplifies the equations. Outside that subject, I think that this equation is not used anymore. Currently it is easier to perform a numerical integration of the movement at the lab frame.--Guentherwagner (talk) 21:32, 17 May 2010 (UTC)

I understand. At the present time I come to the following conclusions :

1/The formula given in the existing text is false. '''(No, it is right) '''

2/The formula I calculated for $$\mathbf\omega$$  in the $$(e)$$ reference frame is right.

3/The formula $$R \mathbf\omega$$ as a function of $$(\alpha ...,\dot\alpha ....)$$ is right but seems not useful : it does not give the coordinates of $$\mathbf\omega$$ in the $$(h)$$ reference frame.

4/I found in "Geometric Algebra for Physicists" (Doran, Lasenby , Cambridge UK) a method which looks like, but perhaps is not quite identical with the one you refer to. It conslsts of rotating back the $$\mathbf\omega$$ vector in a body copy $$\mathbf\omega_C$$ considered in the $$(e)$$ reference frame. $$\mathbf\omega_C$$ is thus an auxiliary vector which permits to write the true angular momentum vector $$L=R \mathbf J\ R^T$$ where J is an inertial tensor function of time only through $$\mathbf\omega_C$$.

5/As a conclusion I think there is no need to try to explicit $$\mathbf\omega $$ in the $$(h)$$ reference system.Chessfan (talk) 13:10, 19 May 2010 (UTC)Chessfan (talk) 14:21, 28 May 2010 (UTC) Chessfan (talk) 17:31, 11 June 2010 (UTC)

Combining rotations
A big problem that I see in your text is when you say "Combining successive Euler rotations" and then you describe the normal rotation composition. When you combine complex operations like precessions and nutations, you have to operate in different frames, changing the basis of your matrices to the mobile frame and back. I think you really just mean "combining successive rotations".--Guentherwagner (talk) 12:45, 19 May 2010 (UTC)

I modify my text, but if you take a close look at the $$(t)_e$$ formula you will see that we indeed have there the two interpretations of the global rotation. Now what shall we do to progress ? I am not able to adjust the editing. Chessfan (talk) 13:31, 19 May 2010 (UTC)

In GA you would not have to worry about the successive frames ; you define coordinate free the successive rotation axes, combine the rotations by algebraic operations and simply calculate the coordinates that interest you in a final operation. Try it and you will be convinced ! Chessfan (talk) 14:07, 19 May 2010 (UTC)


 * Of course I know that composing rotations you can reach any target frame!! but this article is about Euler angles, not rotations in general. The section that speaks about "Relationship with physical motions" is nearly an off-topic. To extend it to include GA (I suppose you mean geometric algebra) is clearly outside the scope of the article. I would propose that you put a comment about GA in the article and you expand the subject in the article Rotation representation (mathematics). It is a much better place. —Preceding unsigned comment added by Guentherwagner (talk • contribs) 21:55, 19 May 2010 (UTC)

Do not worry ! I think I have done my job, which was to help detecting and correcting some errors or at least a lack of definitions. I leave it up to you and other experienced editors to finish the reconstruction of the article. Feel free to correct my text. Concerning GA I will perhaps try to follow your advice.Chessfan (talk) 13:46, 20 May 2010 (UTC)

Geometric Algebra
In articles like Euler angles or Rotation representation (mathematics) the quaternion tool is often cited, much more than geometric algebra. And often the word quaternion is employed for designing the GA rotors, which introduces a prejudiciable confusion. One must know :

1/ that the quaternions are an even subalgebra of GA ;

2/ that GA is much easier to use, mainly because it is more coordinate-free than quaternions;

3/ that GA is easier to interprete because what Hamilton called vectors are indeed bivectors.

Let us show briefly how to express a rotation in GA. The principal tool is the rotor :

(1) ..... ......... ...........................$$ \mathbf\R=[\cos(\theta/2)-I u \sin(\theta/2)] $$

where :

$$\mathbf\theta =$$angle of rotation

$$\mathbf(u) =$$rotation axis (unitary vector)

$$\mathbf(I)=$$pseudoscalar (trivector in $$\mathbb{R}^3)$$

To rotate a vector $$x$$ to position $$y$$ we must write :

(2).............................................$$y=R x \tilde{R}$$

where :

(3).............................................$$\tilde{R}=[\cos(\theta/2)+I u \sin(\theta/2)]$$

is the reverse of the rotor $$\mathbf\R$$.

In the way we have written it the rotation will operate counter clockwise around $$\mathbf(u) $$.

We note immediately one of the main advantages of the rotor versus matrix algebra : we know without auxiliary calculation the rotation axis and angle. So when composing the successive rotations of Euler angles there is no ambiguity about their signification. For instance in the (z,x,z) convention in the above cited article, we get (with (e), (f), (g) being the successive Euler reference frames) :

(4)..............................................$$\mathbf\R=R_\gamma R_\beta R\alpha$$.....................$$\tilde{R}=\tilde{R}_\alpha \tilde{R}_\beta \tilde{R}_\gamma$$

.................................................$$ \mathbf\R_\alpha=[\cos(\alpha/2)-I e_3 \sin(\alpha/2)] $$

.................................................$$ \mathbf\R_\beta=[\cos(\beta/2)-I f_1 \sin(\beta/2)] $$

.................................................$$ \mathbf\R_\gamma=[\cos(\gamma/2)-I g_3 \sin(\gamma/2)] $$

where :

(5)..............................................$$ f_1=R_\alpha e_1 \tilde{R}_\alpha $$

(6)..............................................$$ g_3=R_\beta R_\alpha e_3 \tilde{R}_\alpha \tilde{R}_\beta $$

What follows might seem a bit complicated, but is in fact very easy even for a beginner in GA. It is just a funny game with anti-commuting orthonormal vectors. Let us define :

(7)..............................................$$ \mathbf\R_\alpha=[\cos(\alpha/2)-I e_3 \sin(\alpha/2)]=B_\alpha $$

...............................................................$$\mathbf[\cos(\beta/2)-I e_1 \sin(\beta/2)]=B_\beta $$

...............................................................$$\mathbf[\cos(\gamma/2)-I e_3 \sin(\gamma/2)]=B_\gamma $$

Then we get :

(8)..............................................$$\mathbf\R_\beta R\alpha=(\cos\beta-I R_\alpha e_1 \tilde{R}_\alpha\sin\beta)R_\alpha=R_\alpha(\cos\beta-I e_1 \sin\beta)\tilde{R}_\alpha R_\alpha=B_\alpha B_\beta$$

We leave it to the reader to demonstrate samewise :

(9)..............................................$$\mathbf\R_\gamma R_\beta\ R\alpha=B_\alpha B_\beta B_\gamma=R$$

Of course :

(10)............................................$$\tilde{R}=\tilde{R}_\alpha \tilde{R}_\beta \tilde{R}_\gamma=\tilde{B}_\gamma\tilde{B}_\beta\tilde{B}_\alpha$$

Thus we have demonstrated the double interpretation of the Euler rotation, the first one around combined non orthonormal axis $$(e_3,f_1,g_3)$$, the second around the fixed orthonormal axis $$(e_3,e_1,e_3)$$. Both correspond to the same active rotation.

If $$\theta$$ is the global rotation angle and $$u$$ the axis unitary vector we find :

(11)..$$\mathbf\cos(\theta/2)=\cos(\alpha/2)\cos(\beta/2)\cos(\gamma/2)-\sin(\alpha/2)\cos(\beta/2)\sin(\gamma/2)$$

(12)..$$\mathbf{u}=[1/\sin(\theta/2)](e_1[\cos(\alpha/2)\sin(\beta/2)\cos(\gamma/2)+\sin(\alpha/2)\sin(\beta/2)\sin(\gamma/2)]$$

.........................................$$\mathbf+e_2[\sin(\alpha/2)\sin(\beta/2)\cos(\gamma/2)-\cos(\alpha/2)\sin(\beta/2)\sin(\gamma/2)]$$

.........................................$$\mathbf+e_3[\cos(\alpha/2)\cos(\beta/2)\sin(\gamma/2)+\sin(\alpha/2)\cos(\beta/2)\cos(\gamma/2)])$$

That is :

(13)..$$\mathbf\cos(\theta/2)=\cos(\beta/2)\cos[(\alpha/2)+(\gamma/2)]$$

(14)..$$\mathbf{u}=[1/\sin(\theta/2)](e_1\sin(\beta/2)\cos[(\alpha/2)-(\gamma/2)]+e_2\sin(\beta/2)\sin[(\alpha/2)-(\gamma/2)]+e_3\cos(\beta/2)\sin[(\alpha/2)+(\gamma/2)])$$

One cannot fail to note that the expression of the global rotor is much more symmetric than the equivalent rotation matrix, and quaternions, which seems to confirm the idea that beauty and efficiency often coincide ...

Chessfan (talk) 22:23, 24 May 2010 (UTC)

Sorry, I realise I am losing my time here ; thank you PAR and Guentherwagner ; perhaps JohnBlackburne can help you finishing the new article Good bye. Chessfan (talk) 14:14, 29 May 2010 (UTC)

Additional Remarks
Well, cooling down about being censored whenever I mention GA, and as I hate leaving unfinished jobs, I will assemble here some additional remarks which might perhaps be useful to people reconstructing the Euler angles article.

Euler rotations
As I already asked for, that whole subsection should be suppressed, as almost everything is false (wrong order) or at least ambiguous. Starting with my definitions (see new text) we have :


 * $$R =A(\alpha)B(\beta)C(\gamma)$$


 * $$A(\alpha+\delta\alpha)=A(\delta\alpha) A(\alpha)=(I+\delta\alpha J_3)A(\alpha)$$


 * $$B(\beta+\delta\beta)=B(\delta\beta) B(\beta)=(I+\delta\beta J_1)B(\beta)$$


 * $$C(\gamma+\delta\gamma)=C(\delta\gamma) C(\gamma)=(I+\delta\gamma J_3)C(\gamma)$$

where $$(J_3, J_1)$$ are the rotation generator matrices associated with the third and first axis.


 * $$\delta{R}=\delta(A)BC + A\delta(B)C + AB\delta(C)$$=$$\delta\alpha J_3ABC+\delta\beta A J_1BC+\delta\gamma ABJ_3C$$


 * $$\delta{R}=(\delta\alpha J_3+\delta\beta AJ_1\tilde{A}+\delta\gamma ABJ_3\tilde{B}\tilde{A})R$$


 * $$(t)_e=ABC(u)_e=R(u)_e$$


 * $$(\delta{t})_e=(\delta\alpha J_3+\delta\beta AJ_1\tilde{A}+\delta\gamma ABJ_3\tilde{B}\tilde{A})R (u)_e=(\delta\alpha J_3+\delta\beta AJ_1\tilde{A}+\delta\gamma ABJ_3\tilde{B}\tilde{A})(t)_e$$

We recognize in the term between parentheses the rotation tensor. As by construction this is a first order developpement in $$\delta\alpha$$,$$\delta\beta$$,$$\delta\gamma$$ there is no commutativity problem for small variations. I do not understand what the first editor had in his mind ?

Nevertheless this formula is interesting as we derive from it the velocity pseudovector :


 * $$\dot(t)_e=(\dot\alpha J_3+\dot\beta AJ_1\tilde{A}+\dot\gamma ABJ_3\tilde{B}\tilde{A})(t)_e$$

which is the matrix expression for :


 * $$\dot t=\omega \times t=(\dot\alpha e_3+\dot\beta f_1+\dot\gamma g_3) \times t$$

Chessfan (talk) 14:50, 7 June 2010 (UTC)

Extrinsic Intrinsic
I wrote rotation sections in matrix notation, trying to be non-ambiguous, notably by indexing the matrices with the reference system in which they are supposed to operate. Of course that will probably be called non-encyclopaedical writing ... Well, whatever you decide to do with my work, the existing text for intrinsic and extrinsic rotations should be suppressed or at least carefully rewritten.

The formula :


 * Z&Prime;(&alpha;) o  X&prime;(&beta;)  o  Z(&gamma;) = [ (X&prime;(&beta;)z(&gamma;))  o  z(&alpha;)  o  (X&prime;(&beta;)z(&gamma;))&minus;1 ]  o  X&prime;(&beta;)  o  z(&gamma;)
 * = [ {z(&gamma;)x(&beta;)z(&minus;&gamma;) z(&gamma;)} o  z(&alpha;)  o  {z(&minus;&gamma;) z(&gamma;)x(&minus;&beta;)z(&minus;&gamma;)} ]  o  [ z(&gamma;)x(&beta;)z(&minus;&gamma;) ]  o  z(&gamma;)
 * = z(&gamma;)x(&beta;)z(&alpha;)x(&minus;&beta;)x(&beta;) = z(&gamma;)x(&beta;)z(&alpha;).

looks like my matrix formula, but apart of again the wrong order, is very difficult to understand. It contains operator notation, but seems mixed with matrix multiplication, and then you do not know in which reference system they operate. And the last paragraph about left and right multiplication adds to the confusion. Chessfan (talk) 17:24, 4 June 2010 (UTC)

Velocity vector again, with Geometric Algebra
In the above cited example ( $$zxz$$ convention) I cited already the relation :


 * $$\omega=(\dot\gamma\sin\beta\sin\alpha+\dot\beta\cos\alpha){\mathbf e_1}

+(-\dot\gamma\sin\beta\cos\alpha+\dot\beta\sin\alpha){\mathbf e_2} +(\dot\gamma\cos\beta+\dot\alpha){\mathbf e_3}$$

It is not so easy but we would like to express the velocity vector in the final (h) reference. I will work unencyclopaedically (!) with Geometric Algebra and explain demonstrations.

We start with the obvious relation :


 * $$\omega=\dot\alpha e_3+\dot\beta f_2 +\dot\gamma g_3$$

We must express the vectors of that mixed reference frame as functions of the (h) vectors. That necessitates to work with rotors only related to those vectors. We know already that :


 * $$\mathbf\R_\gamma R_\beta\ R_\alpha=B_\alpha B_\beta B_\gamma=R$$

Now we define :


 * $$ \mathbf\Complex_\alpha=[\cos(\alpha/2)-I h_3 \sin(\alpha/2)] $$


 * $$ \mathbf\Complex_\beta=[\cos(\beta/2)-I h_1 \sin(\beta/2)] $$


 * $$ \mathbf\Complex_\gamma=[\cos(\gamma/2)-I h_3 \sin(\gamma/2)] $$

It is easy to show that :


 * $${B}_\alpha=\tilde{R}C_\alpha R \qquad {B}_\beta=\tilde{R}C_\beta R \qquad {B}_\gamma=\tilde{R}C_\gamma R$$

Then we get :


 * $$\mathbf{C}_\alpha C_\beta C_\gamma=R$$

Now we can write :


 * $$f_1=\tilde{R_\beta}\tilde{R_\gamma}h_1 R_\gamma R_\beta=\tilde{C_\gamma}\tilde{C_\beta}h_1 C_\beta C_\gamma \qquad e_3=\tilde{R} h_3 R\qquad g_3=h_3$$

All rotors can now be expressed as functions of only the (h) vectors. Well I will spare you the details. We find :


 * $$\mathbf\omega=(\dot\alpha \sin\beta \sin\gamma+\dot\beta \cos\gamma)h_1+(\dot\alpha \sin\beta \cos\gamma-\dot\beta \sin\gamma) h_2+(\dot\alpha \cos\beta+\dot\gamma) h_3$$

'''Important : That is the same result as the formula given in the existing article. '''

Chessfan (talk) 07:55, 13 June 2010 (UTC)

We find the same results in http://audiophile.tam.cornell.edu/~als93/Publications/SchwabMeijaard2006.pdf, formulas (14) and (16). Chessfan (talk) 19:48, 25 June 2010 (UTC)

Passive transformations
In the user workpage I studied an active transformation with Euler angles. Let us now look at a passive transformation. The easiest way is to operate with tensorial calculus. We maintain the body fixed and rotate with the successive Euler angles $$\alpha,\beta,\gamma$$ an initial reference frame (h) towards (g), then (f) , then (e). We get :


 * $$g_j=h_i \alpha^i_j \qquad f_k=g_j \beta^j_k \qquad e_m=f_k \gamma^k_m$$


 * $$e_m=\gamma^k_m\beta^j_k\alpha^i_jh_i$$

A vector $$x$$ noted $$\xi$$ in (e) will be written :


 * $$x=\xi^m e_m=\xi^m\gamma^k_m\beta^j_k\alpha^i_jh_i=x^ih_i$$

which gives :


 * $$x^i=\alpha^i_j\beta^j_k\gamma^k_m\xi^m$$

Thus in matrix notation we have (licol rule) :


 * $$A B C (x)_e=(x)_h$$


 * $$(x)_e=C^T B^T A^T (x)_h$$


 * $$R=C^T B^T A^T$$

With simple examples it is easy to show that in the zxz convention the $$C^T, B^T, A^T$$ are precisely the transposed matrices of the $$C,B,A$$ matrices we used in the active transformation. Now multiplying those matrices we find :


 * $$\mathbf{R} = \begin{bmatrix}

c_\alpha c_\gamma - c_\beta s_\alpha s_\gamma & c_\gamma s_\alpha + c_\alpha c_\beta s_\gamma & s_\beta s_\gamma \\ -c_\beta c_\gamma s_\alpha - c_\alpha s_\gamma & c_\alpha c_\beta c_\gamma - s_\alpha s_\gamma & c_\gamma s_\beta \\ s_\alpha s_\beta & -c_\alpha s_\beta & c_\beta \end{bmatrix}$$

We could have obtained that result much quicker by taking the formula which I have demonstrated for an active transformation, changing signs of all angles and then switching $$\alpha$$ and $$\gamma$$. But by establishing both formulas independently we check their rightness.

It is of outmost importance in any article citing rotation matrices to specify if it concerns an active or a passive transformation. That distinction should not be confused with the double interpretation, extrinsic or intrinsic, of a given Euler rotation.

Chessfan (talk) 22:29, 4 July 2010 (UTC)

Are Euler rotations commutative ?
I write this section for people ready to enter deeply in the maths. I will do it mainly with Geometric Algebra, which is better adapted than matrix algebra for dealing with rotations.

I had a difficult discussion with Guentherwagner on the point of determining if successive Euler rotations can be considered being well defined mathematical operators without taking in account the true position of the successive rotation axes, or what is the same thing without attaching each rotation matrix to a specific frame. If the answer is yes can we say that they commute, despite the fact that we know that rotations genrally do not commute ?

I tried to understand and perhaps modify the matricial quasi-demonstration in the text I critizised. The difficulty which renders that problem confusing is the fact that the simple matrices used give only a general indication on the rotation axis. If for instance you work with a simple matrix characterizing a rotation around OZ (with the number one in the lower right corner), you can apply it in any frame and thus choose arbitrarily your Z axis ... ! By the way, that explains the inverse order of the matrices in composing a combined Euler angle rotation. So I was perturbed by the fact that of course in matricial expressions we have $$N P \ne P N$$. But as the matrices on both sides are in fact applied to different frames it gives perhaps in operator notation $$N_1 P_2=P_1 N_2$$. That could be, but not so easily, expressed in matrix formulation. A contrario Geometric Algebra is particularly adapted to that task, and will perhaps reserve us a surprise.

The whole question can be studied by beginning with a fixed reference frame $$(e_1,e_2,e_3)$$ and doing first a precession $$\alpha$$ around $$e_3$$, followed by a nutation $$\beta$$ around $$e'_1$$, second a nutation $$\beta$$ around $$e_1$$ followed by a precession $$\alpha$$ around $$e'_3$$. Then we compare the two global rotors and their action on well chosen vectors. It is important to note that here the different rotors are mathematically rigorously defined ; so we will have an unambiguous answer to the commutativity question. There is no need to represent the rotations which occured before the ones we study (we take them for 0).

In GA the rotation operator is the rotor, in formulas like :

$$(1) \qquad v=Ru \tilde{R}\qquad \qquad R=\cos (\theta/2)- I w \sin(\theta/2) $$

When writing Euler rotations, a precession followed by a nutation (active rotations) will be noted :

$$(2) \qquad R=N'P=(\cos (\beta/2)-I e'_1 \sin(\beta/2))(\cos (\alpha/2)-I e_3 \sin(\alpha/2))$$

with

$$(3) \qquad e'_1=P e_1 \tilde{P}$$

Thus, after some algebra :

$$(4) \qquad N'P=(\cos (\alpha/2)-I e_3 \sin(\alpha/2))(\cos (\beta/2)-I e_1 \sin(\beta/2))$$

Now a nutation followed by a precession will be noted :

$$(5) \qquad R=P'N=(\cos (\alpha/2)-I e'_3 \sin(\alpha/2))(\cos (\beta/2)-I e_1 \sin(\beta/2))$$

with

$$(6) \qquad e'_3=N e_3 \tilde{N}$$

Thus :

$$(7) \qquad P'N=(\cos (\beta/2)-I e_1 \sin(\beta/2))(\cos (\alpha/2)-I e_3 \sin(\alpha/2))$$

It is obvious that the expressions (4) and (7) are not equal, as they have now the structure of two successive rotations in a fixed frame. We verify by developping them :

$$(8) \qquad N'P=\cos (\alpha/2)\cos (\beta/2)- I e_1 cos (\alpha/2)\sin(\beta/2)-I e_2 \sin(\alpha/2)\sin(\beta/2)-I e_3 \cos (\beta/2)\sin(\alpha/2)$$

$$(9) \qquad P'N=\cos (\alpha/2)\cos (\beta/2)- I e_1 cos (\alpha/2)\sin(\beta/2)+I e_2 \sin(\alpha/2)\sin(\beta/2)-I e_3 \cos (\beta/2)\sin(\alpha/2)$$

$$(10) \qquad N'P-P'N=-2 I e_2 \sin(\alpha/2)\sin(\beta/2) $$

By comparing (1) and (8), (9), we see that the two combined rotations have the same angle, given by :

$$(11) \qquad \cos (\theta/2)=\cos (\alpha/2)\cos (\beta/2)$$

but not quite the same axis given by :

$$(12) \qquad w=(\sin(\theta/2))^{-1}( e_1 cos (\alpha/2)\sin(\beta/2)\pm e_2 \sin(\alpha/2)\sin(\beta/2)+ e_3 \cos (\beta/2)\sin(\alpha/2))$$

The nutation angle is often small ; then we can write, neglecting the second order terms in $$\beta$$ :

$$(13) \qquad \theta=\alpha$$

$$(14) \qquad w=(\sin(\alpha/2))^{-1}( e_1 cos (\alpha/2)(\beta/2)\pm e_2 \sin(\alpha/2)(\beta/2)+ e_3 \sin(\alpha/2))$$

The global rotation axis remains near to the precession axis and the global angle is almost equal to the precession angle.

We can also calculate the difference between two vectors rotated from an initial vector. To be simple let us take for example $$e_1$$. We get after all calculations done :

$$(15) \qquad e'_1=N'P e_1 \tilde{P} \tilde{N'}=e_1 \cos\alpha+e_2 \sin\alpha$$

$$(16) \qquad e''_1=P'N e_1 \tilde{N} \tilde{P'}=e_1 \cos\alpha+e_2 \sin\alpha\cos\beta+e_3 \sin\alpha\sin\beta$$

Thus if $$\beta$$ is small the difference is :

$$(17) \qquad e''_1-e'_1= e_3 \beta \sin\alpha$$

Of course the fact that $$e'_1$$ is in the $$(e_1,e_2)$$ plane, and $$e''_1$$ is not, can be established by simple geometric means !

Our conclusion is that finite precessions and nutations, rigorously defined do not commute.

Addendum : The same demonstration is in fact stunningly simple with matrix algebra ; I cannot understand why I did not see that immediately.

The composed matrix for precession followed by nutation can be written $$N'P$$ which is equal to $$PN$$ in extrinsic interpretation (see "combining successive rotations" User:Chessfan/work1. The same matrix for nutation followed by precession can be written $$P'N$$ which is equal to $$NP$$ in extrinsic interpretation. But we know that $$PN \ne NP$$ ; thus $$N'P \ne P'N$$. End of the demonstration.

Chessfan (talk) 12:05, 2 October 2010 (UTC)

A general method for interpreting simple and composed rotation matrices.
It seems that a vast majority of authors and professors forget, when writing about rotation matrices, to mention a very basic fact : a rotation matrix is or should be a mathematical tool which transforms not vectors but coordinates of vectors. Secondly they often forget to mention if they speak about active rotations or passive ones.

An active rotation is when rotating an e-frame to an f-frame you rotate simultaneously a vector attached to the e-frame (that is a rigid body rotation), and you want to know the coordinates of the rotated vector in the initial frame. Then you get following relations in tensorial notation :

$$(1)\qquad f_j=\alpha^i_j e_i \qquad x=x^i e_i \qquad \xi=x^j f_j=\xi^i e_i$$

thus :

$$(2)\qquad \xi^i=\alpha^i_j x^j$$

Now we can write in matricial language :

$$(3)\qquad (\xi)=A (x) \qquad A=|\alpha^i_j| $$

where $$(i,j)$$ are respectively the row and the column indices, and $$(x), (\xi)$$ are 1-column matrices. It is the relation (3) and not the first relation (1) we must use to define the rotation matrix. One finds out easily by picking out the individual $$f_j$$ vectors that the j-th column of the matrix is constituted by the coordinates of $$f_j$$ in the e-frame. One has the impression that many authors write down horizontally (of course ...) the first relation (1), and then conclude falsely that the transposed matrix $$A^T$$ is the valid rotation matrix !

A simple example is given by an active rotation around the Z-axis by an angle $$\alpha$$ positively counted in counterclock direction. The rotation matrix is :


 * $$A=\begin{bmatrix}

\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

What about a passive rotation ? We start with the same above mentioned rotation of the frame $$(e)$$ ⇒$$(f)$$. Now we do not rotate the vector $$x$$ attached to $$(e)$$, but we want to know the coordinates of $$x$$ in $$(f)$$. We can write :

$$(4) \qquad x=x^i e_i=\xi ^j f_j=\xi^j \alpha^i_j e_i$$

$$(5) \qquad x^i= \alpha^i_j \xi^j \qquad (x)=A (\xi)$$

$$(5) \qquad (\xi)=A^{-1}(x)= A^T(x)$$


 * $$A^T=\begin{bmatrix}

\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

We are ready now to tacle the confusing question of Euler angles and rotations.

Let us suppose somebody tells us a global active rotation matrix can be represented by the following matrix product :

$$(6) \qquad R=A B C$$

That means that the coordinates in the initial frame of a vector $$(x)$$ which we rotate by $$R$$ will be transformed to $$(\xi)$$ in the same initial frame by the matricial relation :

$$(7) \qquad (\xi)= R (x)= A B C (x)$$

That means, or should mean in somebodys mind, as the $$A,B,C$$ matrices cannot be anything else than elemental matrices, that we have an extrinsic rotation executed by successive rotations around the initial frame basis vectors in the order C, then B , then A. We have no choice, as each rotation axis is predetermined by the structure of the matrix ! And finally that means that the same rotation matrix can be interpreted as composed by intrinsic  rotations around moved axis, but in the inverse order A ,B',C". Of course the $$B', C''$$, matrices are no more elemental, and will not appear explicitly. Matricially that can be written as follows :

$$(8) \qquad (\xi)= R (x)= A B C (x)=(ABCB^TA^T)(ABA^T)A (x)=C'' B' A (x)$$

That begins to look quite simple. Now we are able to verify what is told in the books !

What if we have to interprete passive Euler rotations ? To be able to compare with the active rotations we choose to use the same A, B, C matrices as before -- that means the rotation (order and angles) of the successive frames are unchanged --, but of course now for each elemental rotation we have a relation of type :

$$(9) \qquad (\xi)=A^T (x)$$

To be sure not to make errors it will be save to write explicitly the rotation of the frames :

$$(10) \qquad h_m= \gamma^k_m \beta^j_k \alpha^i_j e_i$$

$$'11) \qquad x^i e_i=\xi^m h_m=\xi^m \gamma^k_m \beta^j_k \alpha^i_j e_i$$

$$(12) \qquad x^i=\xi^m \gamma^k_m \beta^j_k \alpha^i_j=\alpha^i_j\beta^j_k\gamma^k_m \xi^m$$

The reversing of the order is necessary to comply to matrix multiplication rules (licol ...) :

$$(13) \qquad (x)= A B C (\xi)$$

$$(14) \qquad (\xi)=C^T B^T A^T (x)= (A B C)^T (x)= R^T (x)$$

So we happily find the same transposition relation between the global rotation matrices as between the simple matrices when we interpret the same physical rotation in alternatively active or passive mode.

But we have still not told the end of the story. Indeed it is important to say that in (14) we have in $$(\xi)$$ the coordinates of the $$x$$ vector -- attached to (e)--, observed in the (h) frame, which correspond to a passive rotation. But we could also interprete (14) in a quite different manner. By comparing with (8) we could imagine a composed rotation from (e) to a frame (h')≠ (h) composed by intrinsic rotations with the successive order and values $$(-\gamma), (-\beta), (-\alpha)$$. Then $$(\xi)$$ would represent the coordinates in (e) of the negatively rotated vector $$x$$ (active intrinsic rotation). The same vector $$\xi$$ can also be obtained by rotating extrinsicly the vector $$x$$ around the basis vectors of (e) successively in the order and with values $$(-\alpha),(-\beta),(-\gamma)$$.

When working with Euler angles be precise, look at the math, and do not spare the indices. Most of the obscurities and misunderstandings arise from the fact that the authors do not follow that advice.

Chessfan (talk) 16:29, 27 January 2012 (UTC)

Matrix orientation (new proposal).
Using the equivalence between Euler angles and rotation composition, it is possible to change to and from matrix convention. First, active rotations and the right handed rule for the positive sign of the angles are assumed. As explained above each global rotation matrix can be interpreted as a composition of intrinsic rotations, that is around the axes of the moving body (the moving reference system), or as a composition of extrinsic rotations around the fixed world axes. We choose to name the different conventions by privileging the latter, for a simple and obvious reason : in a matrix composition like

$$(1) \qquad R= A B C$$

the composing matrices are elemental, and thus pick out automatically successively different fixed world axes in the order C,B,A. We, of course, do not use the complicated intrinsic expression C"B'A to calculate $$R$$.

To suppress all existing ambiguities and facilitate the understanding between different professional users we introduce a specific denomination principle more detailed than the existing ones. We introduce following rules :

1/ When naming a global rotation we employ the same names in the same order than the matrix formula used to calculate the global matrix. If $$R=Z_\alpha X_\beta Z_\gamma$$ we name it $$Z_\alpha X_\beta Z_\gamma$$. The consequence of that is that the name illustrates automatically the extrinsic interpretation.

2/ For each matrix we designate explicitly the axis, which we index with the angles (we abbreviate $$\theta_1, ...$$ with $$1,2,3)$$. That may seem redundant but in fact is not, because there is no valid reason to associate an alphabetic or numeric order to the angles, when the order of rotations is already fixed by rule 1.

3/ We recall something that should be obvious, that is the composing and composed matrices are supposed to act on the coordinates of vectors defined in the initial fixed world axis and give as a result the coordinates in the same system of a rotated vector, rotation primarily interpreted extrinsicly, but to which we can give globally an equivalent intrinsic interpretation.

With those rules we get the following matrix table :


 * {| class="wikitable" style="background-color:white;font-weight:bold"

!$$X_1Z_2X_3$$ c_2 & - c_3 s_2 & s_2 s_3 \\ c_1 s_2 & c_1 c_2 c_3 - s_1 s_3 & - c_3 s_1 - c_1 c_2 s_3 \\ s_1 s_2 & c_1 s_3 + c_2 c_3 s_1 & c_1 c_3 - c_2 s_1 s_3 \end{bmatrix}$$ !$$X_1Z_2Y_3$$ c_2 c_3 & - s_2 & c_2 s_3 \\ s_1 s_3 + c_1 c_3 s_2 & c_1 c_2 & c_1 s_2 s_3 - c_3 s_1 \\ c_3 s_1 s_2 - c_1 s_3 & c_2 s_1 & c_1 c_3 + s_1 s_2 s_3 \end{bmatrix}$$ !$$X_1Y_2X_3$$ c_2 & s_2 s_3 & c_3 s_2 \\ s_1 s_2 & c_1 c_3 - c_2 s_1 s_3 & - c_1 s_3 - c_2 c_3 s_1 \\ - c_1 s_2 & c_3 s_1 + c_1 c_2 s_3 & c_1 c_2 c_3 - s_1 s_3 \end{bmatrix}$$ !$$X_1Y_2Z_3$$ c_2 c_3 & - c_2 s_3 & s_2 \\ c_1 s_3 + c_3 s_1 s_2 & c_1 c_3 - s_1 s_2 s_3 & - c_2 s_1 \\ s_1 s_3 - c_1 c_3 s_2 & c_3 s_1 + c_1 s_2 s_3 & c_1 c_2 \end{bmatrix}$$ !$$Y_1X_2Y_3$$ c_1 c_3 - c_2 s_1 s_3 & s_1 s_2 & c_1 s_3 + c_2 c_3 s_1 \\ s_2 s_3 & c_2 & - c_3 s_2 \\ - c_3 s_1 - c_1 c_2 s_3 & c_1 s_2 & c_1 c_2 c_3 - s_1 s_3 \end{bmatrix}$$ !$$Y_1X_2Z_3$$ c_1 c_3 + s_1 s_2 s_3 & c_3 s_1 s_2 - c_1 s_3 & c_2 s_1 \\ c_2 s_3 & c_2 c_3 & - s_2 \\ c_1 s_2 s_3 - c_3 s_1 & s_1 s_3 + c_1 c_3 s_2 & c_1 c_2 \end{bmatrix}$$ !$$Y_1Z_2Y_3$$ c_1 c_2 c_3 - s_1 s_3 & - c_1 s_2 & c_3 s_1 + c_1 c_2 s_3 \\ c_3 s_2 & c_2 & s_2 s_3 \\ - c_1 s_3 - c_2 c_3 s_1 & s_1 s_2 & c_1 c_3 - c_2 s_1 s_3 \end{bmatrix}$$ !$$Y_1Z_2X_3$$ c_1 c_2 & s_1 s_3 - c_1 c_3 s_2 & c_3 s_1 + c_1 s_2 s_3 \\ s_2 & c_2 c_3 & - c_2 s_3 \\ - c_2 s_1 & c_1 s_3 + c_3 s_1 s_2 & c_1 c_3 - s_1 s_2 s_3 \end{bmatrix}$$ !$$Z_1Y_2Z_3$$ c_1 c_2 c_3 - s_1 s_3 & - c_3 s_1 - c_1 c_2 s_3 & c_1 s_2 \\ c_1 s_3 + c_2 c_3 s_1 & c_1 c_3 - c_2 s_1 s_3 & s_1 s_2 \\ - c_3 s_2 & s_2 s_3 & c_2 \end{bmatrix}$$ !$$Z_1Y_2X_3$$ c_1 c_2 & c_1 s_2 s_3 - c_3 s_1 & s_1 s_3 + c_1 c_3 s_2 \\ c_2 s_1 & c_1 c_3 + s_1 s_2 s_3 & c_3 s_1 s_2 - c_1 s_3 \\ - s_2 & c_2 s_3 & c_2 c_3 \end{bmatrix}$$ !$$Z_1X_2Z_3$$ c_1 c_3 - c_2 s_1 s_3 & - c_1 s_3 - c_2 c_3 s_1 & s_1 s_2 \\ c_3 s_1 + c_1 c_2 s_3 & c_1 c_2 c_3 - s_1 s_3 & - c_1 s_2 \\ s_2 s_3 & c_3 s_2 & c_2 \end{bmatrix}$$ !$$Z_1X_2Y_3$$ c_1 c_3 - s_1 s_2 s_3 & - c_2 s_1 & c_1 s_3 + c_3 s_1 s_2 \\ c_3 s_1 + c_1 s_2 s_3 & c_1 c_2 & s_1 s_3 - c_1 c_3 s_2 \\ - c_2 s_3 & s_2 & c_2 c_3 \end{bmatrix}$$
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * }

When we substitute passive rotations to the active ones we have only to transpose the matrix table. Then each matrix transforms the initial coordinates of a vector remaining fixed to the coordinates of the same vector measured in the globally rotated reference system (same rotation axis, same angles).

When we want to establish a matrix table for left-handed sign convention, we need only change the sign of all the sinuses in the tables.

Chessfan (talk) 18:25, 25 February 2012 (UTC)

Establishing coherence in rotation articles.
I begin feeling like Frankenstein's monster roaming through the fields of rotation articles. Wherever I look I find errors or at least false interpretations, ambiguities, lack of understanding between different professional users. Isn't the idea of establishing coherence between these articles an almost impossible task ? With that question in mind I wrote the section "A general method for interpreting simple and composed rotation matrices" in http://en.wikipedia.org/wiki/Talk:Euler_angles. You might perhaps read it.

Now take a look here at the subsection "Rotation matrix ↔ Euler angles".

First the editor gives us an example $$(\phi,\theta,\psi)$$ which is unrelated to the composed matrix given later.

Second we are not told that the detailed matrices correspond to passive rotations.

Third we are not told that $$A_Z A_Y A_X$$ corresponds to the extrinsic interpretation of the global matrix. That ambiguity is reflected in the "Euler rotations" section. What can we understand under "the most external matrix rotates the other two ..." ? Matrices are not intrinsic mathematical objects which can be actively rotated ; they take their signification when attached to a reference system. That is why we must write to fully express intrinsic rotations transformed in extrinsic ones :

$$\qquad (R)_e=(Z''_\gamma)_e (Y'_\beta)_e (X_\alpha)_e=(X_\alpha)_g (Y_\beta)_f(Z_\gamma)_e=(X_\alpha)(Y_\beta)(Z_\gamma)$$

where Z" and Y' are no more elemental matrices.

Well, I think I have done my job in the Euler angles main article, and I have no more time for that. Who will undertake the coordination task ?

Chessfan (talk) 23:25, 26 February 2012 (UTC)

"Rotation" of Spinors?
Since a long time I wonder if some physicists don't introduce an artificial scent of mystery in the spinor question by insisting heavily on vague general definitions like: ''Spinors are rotated in a different manner than vectors. They change sign when rotated by a 360 degrees angle; to restore the original spinor one must rotate it by a 720 degrees angle!'' That is particularly the case in the present Wikipedia article, first in the introduction, second in the 2-d and 3-d examples, which as they are now written are deeply misleading.

It should be clear to everyone, that expressions like:

$$ R_\beta R_\alpha u R'_\alpha R'\beta$$

represent the composition of two successive rotations of a vector $$u$$ first by an angle $$\alpha$$, followed by an angle $$\beta$$, but that interpreting $$R_\beta R_\alpha$$ as the rotation of the rotor $$R_\alpha$$ by the rotor $$R_\beta$$ with an angle $$\beta$$ is a double mathematical interpretation error, because:

1/ The acting rotation parameters in $$R_\beta R_\alpha$$ are not $$\beta$$ and $$\alpha$$ but that angles halved.

2/ Interpreting the unitary transformation of $$R_\alpha$$ by $$R_\beta$$ as a rotation has no pure geometrical signification, neither in geometric algebra nor of course in traditional Clifford algebra. (In GA such a significative rotation would be expressed as $$R_\beta R_\alpha R'_\beta$$, where the scalar part and the bivector part would be separately rotated in a very simple manner). Of course what remains true is the change of sign of $$R_\alpha$$ under the unitary transformation with parameter $$\beta/2=\pi$$, but that has nothing mysterious ! Call it $$R_{2\pi}$$ if you want.

Those facts are not small details ; they justify in my opinion a rewriting of the introduction and the examples.

By the way the editors of the examples have not made a clear choice between a traditional Clifford algebra presentation and GA ; that is also obscuring the question, as the fact that in Clifford algebra the so-called rotations operate in a complex SU(2) space is not even mentioned.

Chessfan (talk) 15:19, 1 September 2013 (UTC)

--Chessfan (talk) 17:14, 12 December 2014 (UTC)

Rotation of spinors, again ...
When I wrote "Rotation of spinors" talk, at least I hoped that some obvious errors in the 2-d and 3-d examples could be removed. Since then the article has become more and more abstract, and nothing has been done to correct the errors. How can one write (in GA) that:

$$(1/\sqrt2)(1-\sigma_1 \sigma_2)(a_1+a_2\sigma_1\sigma_2)=(1/\sqrt2)[a_1+a_2+(-a_1+a_2)\sigma_1\sigma_2]$$

is a rotation (45° or whatsoever)ǃǃ

It is true that

$$\exp(-B\pi)R=-R$$

but that is the end of it. unless you restrain yourself to the spin space.

— Preceding unsigned comment added by Chessfan (talk • contribs) 19:39, 11 December 2014 (UTC)

Time dilation ... another explanation.
Time dilation ... another explanation.

Time dilation is since Einstein reported in a rather complicated manner.

We look at the transported clock "A" with the "point de vue" from our fixed clock "B". And of course we find that "A" runs slow.

But we forget that "all the laws of physics are the same in every inertials frame of reference". That means ː if we transport, at constant speed a sister clock "A" of "B" , that sister clock runs exactly at the same rate than "B".

And if we measure with "A" a certain amount of time between the initial position of "A" and its final position, we will find the correct answers (that is with the Lorentz transformations) , with never having thought that "A" runs slow. "A" and "B" have the same tic-tac but "A" has a shorter time to measure.

That interpretation is not only easier to explain, but it simplifies the experimental analysis. We understand how there is in space-time a sort of trading between space and time. --Chessfan (talk) 20:07, 16 May 2016 (UTC)--Chessfan (talk) 06:06, 18 May 2016 (UTC)

(http://www.fermedesetoiles.com/documents/supports/le-paradoxe-des-jumeaux.pdf page34)--Chessfan (talk) 06:42, 18 May 2016 (UTC)

How to simplify our vision of the twin paradox.
Time dilation is since Einstein reported in a rather complicated manner.

We look at the transported clock "A" with the "point de vue" from our fixed clock "B". And of course we find that "A" runs slow.

But we forget that "all the laws of physics are the same in every inertial frame of reference". That means ː if we transport, at constant speed a sister clock "A" of "B" , that sister clock runs exactly at the same rate than "B".

And if we measure, with "A" , a certain amount of time between the initial position of "A" and its final position , we will find the correct answers (that is with the Lorentz transformations) , with never having thought that "A" runs slow. That means we dissipated any confusion the physical clocks and time they measure.

Of course we must also explain what happens in the acceleration - deceleration phases, but it is now well-known that the contribution of these phases are negligible. And also that modern clocks are almost insensible to accelerations ǃ

That interpretation is not only easier to explain, but it simplifies the experimental analysis. We understand how there is in space-time a sort of trading between space and time.

--Chessfan (talk) 13:45, 30 January 2017 (UTC)